Question

1. -/1 points a car is parked on a cliff overlooking the ocean on an incline that makes an angle of 19.0° below the horizontal. the negligent driver leaves the car in neutral, and the emergency brakes are defective. the car rolls from rest down the incline with a constant acceleration of 3.06 m/s² for a distance of 60.0 m to the edge of the cliff, which is 40.0 m above the ocean. (a) find the cars position relative to the base of the cliff when the car lands in the ocean. m (b) find the length of time the car is in the air. s details my notes ask your teacher submit answer

Explanation:

Step1: Find initial velocity along the incline

We use the kinematic - equation $v^{2}=v_{0}^{2}+2ax$. The car starts from rest ($v_{0} = 0$), $a = 3.06\ m/s^{2}$ and $x = 60.0\ m$. So, $v=\sqrt{2ax}=\sqrt{2\times3.06\times60.0}=\sqrt{367.2}\approx19.2\ m/s$.

Step2: Resolve the initial velocity into horizontal and vertical components

The angle of the incline $\theta = 19.0^{\circ}$. The initial vertical velocity $v_{y0}=-v\sin\theta$ and the initial horizontal velocity $v_{x0}=v\cos\theta$. $v_{y0}=- 19.2\sin(19.0^{\circ})\approx - 6.24\ m/s$ and $v_{x0}=19.2\cos(19.0^{\circ})\approx18.1\ m/s$.

Step3: Find the time of flight for part (b)

We use the vertical - motion kinematic equation $y = y_{0}+v_{y0}t+\frac{1}{2}at^{2}$. Here, $y = 0$ (lands in the ocean), $y_{0}=40.0\ m$, $v_{y0}=-6.24\ m/s$ and $a=-9.8\ m/s^{2}$. So, $0 = 40.0-6.24t - 4.9t^{2}$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $4.9t^{2}+6.24t - 40.0 = 0$, where $a = 4.9$, $b = 6.24$ and $c=-40.0$.
\[t=\frac{-6.24\pm\sqrt{(6.24)^{2}-4\times4.9\times(-40.0)}}{2\times4.9}=\frac{-6.24\pm\sqrt{38.94 + 784}}{9.8}=\frac{-6.24\pm\sqrt{822.94}}{9.8}=\frac{-6.24\pm28.7}{9.8}\]
We take the positive root $t=\frac{-6.24 + 28.7}{9.8}=\frac{22.46}{9.8}\approx2.3\ s$.

Step4: Find the horizontal distance for part (a)

We use the horizontal - motion equation $x = v_{x0}t$. We already found $v_{x0}\approx18.1\ m/s$ and $t = 2.3\ s$. So, $x=18.1\times2.3 = 41.63\ m$.

Answer:

(a) $41.6\ m$
(b) $2.3\ s$